848 Appendix II. 2nd moment areas 02 a'2 are obtained on the same sides of the vertical. Now the real value of I will be found by doubling our results, for we have only used half of the section; hence, I - 2(02 + a'2)j;2 and Z - I = z^ + a'^y or generally, Moment of Resistance = /(2nd moment area)jy if the reference distancey has been adopted. In cast-iron beams jVt is always taken, which is less than yc (see p. 435), and the resulting curves are slightly changed, but the construction ex- plained must still be rigidly followed. It will be seen that this method is superior to that at p. 430, though the latter is still left in the text as sometimes convenient. (Seep. 1079.) P. 441. Fixed Beams.—When beams have their ends fixed as in Figs. 399 and 400, the Bm curve may be found graphically, by supposing it the algebraic sum of two moment curves, one caused by the ' action' of the free load, and the other by the ' reaction' (a couple) in the wall itself. These opposing curves must cover exactly equal areas: for, considering the upper fibres of the beam say, the total effect of the load is to shorten them; but, remembering that the total length of the beam is unalterable on account of its fixedness, the reactionary couple must entirely eliminate "the aforesaid compression. The sum of the strains is therefore zero, and because Bm cc / oc b in uniform sectioned beams, the sum of the Bm (average Bm x /) due to free load must equal the sum of Bm due to wall couple. Taking the special cases of Figs. 399 and 400, the first is easily solved, but the second or uniformly loaded beam will be further explained by Fig. 811. Now the Bm curve for a free beam is a parabola, where b d, ~ W7-f-8; and the wall moment is efgh) where eg = ^(bd\ to make areas A and B equal. The final !Bm curve is shewn shaded at c. Lastly, to find contra- flexure points, take any distance x between k and /. t> * W/ ,. W7 /W * r x Bm at x =----D = — - I— x.- -W • - 12 r»2 \2 / 2